x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x + 2)(x - 2)(x^2 + 4)
The partial fraction decomposition will therefore be of the form A/(x + 2) + B/(x - 2) + (Cx + D)/(x^2 + 4).
A/(x + 2)+ B/(x- 2) + (Cx + D)/(x^2 + 4) = 1/((x + 2)(x - 2)(x^2 + 4))
A(x - 2)(x^2 + 4) + B(x + 2)(x^2 + 4) + (Cx + D)(x +2)(x - 2) = 1
A(x^3 - 2x^2 + 4x - 8) + B(x^3 + 2x^2 + 4x + 8) + (Cx + D)(x^2 - 4)= 1
Ax^3- 2Ax^2 + 4Ax - 8A + Bx^3 + 2Bx^2 + 4Bx + 8B + Cx^3 + Dx^2 - 4Cx - 4D = 1
(A + B + C)x^3 + (2B - 2A + D)x^2 + (4A + 4B - 4C)x + (8B - 8A - 4D) = 1
Write a system of equations now:
{(A + B + C = 0),(2B - 2A + D = 0), (4A + 4B - 4C = 0), (8B - 8A - 4D = 1
):}
First of all, the third equation can be simplified to A + B - C = 0. We can get rid of the C variable by adding the first and the third equations, to get 2A + 2B = 0 -> A + B = 0. This means that B = -A. Also, from the second equation, we deduce that D = 2A - 2B.
Substitute into the fourth equation:
8(-A) - 8A - 4(2A - 2(-A)) = 1
-8A - 8A - 8A - 8A = 1
-32A = 1
A = -1/32
Solving for the other variables is now relatively simple. You should get A = -1/32, B = 1/32,C = 0 and D = -1/8 as final answers.
:.The partial fraction decomposition is -1/(32(x + 2)) + 1/(32(x - 2)) - 1/(8(x^2 + 4))
Let's look closely at the integration of int-1/(8(x^2 + 4))dx. This will be of the arctangent form, but first we need to perform a u-substitution.
Let u = x/2. Then du = 1/2dx.
-1/8int1/(u^2 +1) * 1/2du
-1/16int 1/(u^2 + 1)du
This is a standard integral.
-1/16arctanu
-1/16arctan(x/2)
Now deal with the other parts of the problem. The other two terms can be integrated using int1/xdx = ln|x| + C.
int1/(32(x - 2))dx = 1/32ln|x - 2|
int-1/(32(x +2))dx = -1/32ln|x + 2|
Therefore, int1/(x^4 - 16)dx = 1/32ln|x - 2| - 1/32ln|x + 2| - 1/16arctan(x/2) + C
Hopefully this helps!
